C dereferencing a pointer to array? - c

Question

I have some code that looks like this:
int *array[10];
printf("%p\n", array); // Prints an address
printf("%p\n", *array); // Prints a different address
printf("%d\n", **array); // Segmentation Fault 11
printf("%d\n", *array[0]); // Segmentation Fault 11
printf("%d\n", (*array)[0]); // Segmentation Fault 11
Why do I get a segmentation fault? Shouldn't it print the first value in the first array?

Answer 1

To take a closer look, understand the nature of the declaration:
int *array[10];
This declares an array of 10 pointers (currently uninitialized) to ints -- note that this is not the same as a pointer to an array of 10 ints, which would instead be declared int (*array)[10]. Even declared this way though, you'd still need to initialize the pointer with something.
printf("%d\n", array); // Prints an address
This prints an the address of the ages array (the array is converted to a pointer automatically by passing it to printf).
printf("%d\n", *array); // Prints a different address
This uses the same rules to convert the array to a pointer, and then dereferences that pointer. Therefore, you're printing the first value of the array (equivalent to printf("%d\n", ages[0])). However, what you're actually printing here is an address, not an integer (even if it is uninitialized).
printf("%d\n", **array); // Segmentation Fault 11
printf("%d\n", *array[0]); // Segmentation Fault 11
printf("%d\n", (*array)[0]); // Segmentation Fault 11
Each of these are now dereferencing the uninitialized pointer stored in array[0]. They do indeed be refer to an int, but the pointer to that int is whatever the compiler and/or your OS decided to put in there.
Example: Pointer to an array
An example of using a pointer to an array looks like this:
#include <stdio.h>
int main()
{
int array[10] = { 1,2,3,4,5,6,7,8,9,10 };
int (*parr)[10] = &array;
printf("%p\n", parr);
printf("%p\n", *parr);
printf("%d\n", **parr);
printf("%d\n", *parr[0]);
printf("%d\n", (*parr)[0]);
}
Output (addresses vary, obviously)
0x7fff5fbff990
0x7fff5fbff990
1
1
1
The last three all ultimately lead to the same element, but go about it in different ways.

Answer 2

You get a segmentation fault because you are attempting to dereference an uninitialized pointer, which is Undefined Behavior.
This command (and all following):
**ages
Dereferences the array (which decays to a pointer), which you then dereference. You have an array of pointers, not a pointer to an array.

Answer 3

int *array[10];
array is array[10] of pointer to int (note that [] has higher precedence than *). But none of these 10 pointers have been initialised to point to valid location.
Given that,
printf("%d\n", array); // address of array itself (also address of first element)
printf("%d\n", *array); // getting the value of the fisrt array element - UB/unitilialise (currently points to some random location)
printf("%d\n", **array); // dereference that first pointer - UB/segfault as deferencing an initialised pointer
printf("%d\n", *array[0]); // same as *array
printf("%d\n", (*array)[0]); // same as **array

Answer 4

#include <iostream>
using namespace std;
#include <math.h>
#include <string.h>
int main() {
int arr[5] = {1,2,3,4,5};
int *p=arr;
int intgerSize=sizeof(int);
for(int k=0;k<5;k++)
{
cout<<"arr ["<<k<<"] "<<*(p+(k*sizeof(int)/intgerSize));
cout<<" "<<(p+(k*sizeof(int)/intgerSize));
cout<<" "<<p+k<<"\n";
}`enter code here`
return 0;
}
OUTPUT:-
arr [0] 1 0x7ffd180f5800 0x7ffd180f5800
arr [1] 2 0x7ffd180f5804 0x7ffd180f5804
arr [2] 3 0x7ffd180f5808 0x7ffd180f5808
arr [3] 4 0x7ffd180f580c 0x7ffd180f580c
arr [4] 5 0x7ffd180f5810 0x7ffd180f5810